Mensuration - Perimeter, Area, Volume Grade 7 - Edugain · PDF fileGrade 7 Mensuration -...

Grade 7Mensuration - Perimeter, Area, Volume

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Answer the questions

(1) Find area of shaded part (All dimensions are in cm).

(2) A stadium has a circular race track. The outer circumf erence of the track is 541 m and the inner

circumf erence is 365 m. What is the width of the track (Assume π is 22

7 )

(3) If the diagonal of a square is decreased by 15%, then by what percent does the area of thesquare decrease?

Choose correct answer(s) from given choice

(4) Perimeter of a square = ____ x length of a side

a. 4 b. 3

c. 1 d. 2

(5) If diameter of a wheel of a cycle is 42 cm. If wheel makes 50 rotation in one minute, how f ar willcycle travel in 20 minutes? (assume π = 22/7)

a. 1320 meters b. 1188 meters

c. 1056 meters d. 1452 meters

(6) The sides of a rectangle are in the ratio 8:5. If perimeter of the rectangle is 520 cm, what is thelength of rectangle?

a. 160cm b. 150cm

c. 170cm d. 320cm

ID : ae-7-Mensuration-Perimeter-Area-Volume [1]

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(7) Gamal has Dhs 10004670 with which he wants to buy some land. If he wants to buy a piece ofland that is atleast 39 metres wide, and the cost f or land is Dhs 3018 per square metre, thenwhat will be the length of the longest piece of land he can buy?

a. 85 m b. 99 m

c. 91 m d. 80 m

(8) The f loor of a theater is covered by square tiles, all of the same size, laid side by side to f orm arectangle of 175 tiles by 175 tiles. If you draw a line f rom one corner of the theater to theopposite corner, how many tiles will this line intersect?

a. 169 b. 174

c. 178 d. 175

(9) Two neighbours have adjoining houses as shown below. The houses are f enced, with one sideas common. If the cost of f encing is Dhs 16.20 per metre, then what was the total cost of f encing both thehouses?

a. Dhs 2123.60 b. Dhs 2287.40

c. Dhs 2213.50 d. Dhs 2170.80

(10)

If each small square is of size 1cm x 1cm, f ind the perimeter of next shape in the series.

a. 26 cm b. 20 cm

c. 24 cm d. 22 cm



(11) ABCD is a rectangle, and radius of quarter-circle shown in picture is 9.1 m. Find area of theshaded region (All dimensions are in m, and assume π = 22/7).

a. 1222.935 m2 b. 1220.935m2

c. 1225.935m2 d. 1219.935m2

(12) The diagonal of a rectangle is f our t imes the length of its smaller side. What is the ratio of itslength and breadth?

a. √15:1 b. √3:1

c. 4:1 d. 2√2:1

Fill in the blanks

(13) If circumf erence of a circle is 66.6 cm more than its radius, the radius of the circle is

cm. (assume π = 22/7)

(14) If perimeter of f ollowing semi-circular shape is 180 cm, its radius = cm (assume π =

22/7).

(15) Area of rectangle ABCD and AEFG are equal. If AE=36 cm, DG=6 cm and EB=12 cm the length of

AD = cm.

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Answers

(1) 40 cm2

Step 1

The area of rectangle = length × breath cm2.

The area of right angle triangle = Perpendicular × Base

2 cm2

Step 2

If you look at the f igure caref ully, you will notice that there are right angle triangle in arectangle.

theref ore the area of rectangle = 6 × 9 = 54 cm2,

the area of right angle triangle = 4 × 7

2 = 14 cm2,

Step 3

Now the area of shaded part = area of rectangle - area of right angle triangle = 54 - 14 =

40 cm2



(2) 28

Step 1

Lets assume r1 and r2 are the inner and outer radius of the circular race track as shownbelow.

Now the perimeter of the inner circular race track = 2πr1 mand the perimeter of the outer circular race track = 2πr2 m

Step 2

According to question the outer circumf erence of the track is 541 m and the innercircumf erence is 365 m.Theref ore 2πr1 = 365

⇒ r1 = 365

2π

and 2πr2 = 541

⇒ r2 = 541

2π

Step 3

Now the width of the track = r2 - r1

= 541

2π -

365

2π

= 541 - 365

2π

=

176

2 × 22

7

= 176 × 7

2 × 22

= 28 m

Step 4

Theref ore the width of the track is 28 m.



(3) 27.75%

Step 1

Let the length of the diagonal of the square be d. The length of the side of the square will

then be d / √2, and the area of the square will be (d / √2) × (d / √2) = 0.5d2

Step 2

Af ter reducing the length of the diagonal by 15%, the new length of the diagonal will be:

= d - 15

100 d

= 0.85d

Step 3

Hence the new area will be 0.5(0.85d)2 = 0.5 × 0.7225d2.

Step 4

The decrease in area = Old Area - New Area

= 0.5 d2 - 0.5 × 0.7225d2

= 0.5 × (1 - 0.7225) d2

= 0.5 × 0.2775 d2

Step 5

Percentage decrease in area = Decrease in area

Old Area × 100 %

= 0.5 × 0.2775 d2

0.5 d2 × 100 %

= 0.2775 × 100 %= 27.75%

Step 6

Hence, when the diagonal of a square is decreased by 15%, then the area of the squaredecreases by 27.75%.

(4) a. 4



(5) a. 1320 meters

Step 1

Lets assume r and d are the radius and diameter of the wheel respectively, as shownbelow.

Wheel

Step 2

According to question r = 42

2 = 21 cm

The distance traveled by cycle in one rotation = 2πr

= 2 × 22

7 × 21

= 132 cm

Step 3

Since wheel makes 50 rotation in one minute.Theref ore the number of rotations made by wheel in 20 minutes = 50 × 20 = 1000Now the distance traveled by cycle in 20 minutes = 132 × 1000 = 132000 cm

Step 4

Theref ore cycle travel 1320 meters in 20 minutes.



(6) a. 160cm

Step 1

It is given that the ratio of the sides of the rectangle = 8:5Theref ore, we can assume the length and width of the rectangle be 8x and 5x respectively.

Step 2

It is also given that the perimeter of the rectangle is 520 cm.Since, we know that the perimeter of the rectangle = 2(length + width)Theref ore, 2(length + width) = 520or 2(8x + 5x) = 520or 2(13x) = 520or 26x = 520

or x = 520

26 = 20

Step 3

Thus, the length of rectangle = 8x = 8 × 20 = 160 cm

(7) a. 85 m

Step 1

If you read the question caref ully, you will notice that Gamal has Dhs 10004670 and in Dhs3018 he can buy 1 square meter land.

Theref ore in Dhs 10004670 he can buy the piece of land = 1

3018 × 10004670 = 3315

square metre,and hence the area of the piece of land = 3315 square metre

Step 2

The width of the piece of land is 39 metres.Area of the piece of length = length × width ⇒ 3315 = length × 39

⇒ 3315

39 = length

⇒ 85 = length⇒ length = 85 metres

Step 3

Now the width of the land is 39 metres.

(8) d. 175



(9) d. Dhs 2170.80

Step 1

If you look at both the houses caref ully, you will notice that the middle side is common,theref ore we need't to repeat it.The perimeter of both houses with common side = 2( The length of f irst house + Thelength of second house + The width of both houses ) + The width of the middle side= 2(26 + 23 + 12) + 12= 122 + 12= 134 m

Step 2

Since the cost of f encing is Dhs 16.20 per metre.Theref ore the cost of f encing134 m is = 134 × 16.20= 2170.80

Step 3

Now we can say that the cost of f encing both the houses is Dhs 2170.80

(10) c. 24 cm

Step 1

Perimeter is the distance around a two dimensional shape.Perimeter of a square = 4 × length of the side of a square

Step 2

If you look at the given shapes caref ully, you will notice that the perimeter of the f irstshape in the series = 1 × 4 = 4 cmPerimeter of the second shape in the series = 2 × 4 = 8 cm

Theref ore the perimeter of next shape (i.e 6th shape) in the series = 6 × 4 = 24 cm

Step 3

The perimeter of next shape in the series is 24 cm.



(11) a. 1222.935 m2

Step 1

Lets assume r is the radius of quarter-circle. According to question, the radius of quarter-circle shown in picture is 9.1 m.Theref ore r = 9.1 m

Step 2

Area of a quarter-circle = πr2

4

= πr2

4

= 22

7 × (9.1)2

4

= 22 × 9.1 × 9.1

28

= 65.065 m2

Step 3

Area of the shaded region is = area of rectangle ABCD - area of quarter-circle= (AB × BC) - 65.065= (46 × 28) - 65.065= 1288 - 65.065

= 1222.935 m2

Step 4

Area of the shaded region is 1222.935 m2.



(12) a. √15:1

Step 1

Lets assume ABCD is a rectangle.

Step 2

Lets assume b and l is the breath(smaller side) and length of the rectangle respectively.Since the diagonal of a rectangle is f our t imes the length of its smaller side.theref ore the length of the diagonal of the rectangle = 4b cm.

Step 3

If you look at the rectangle ABCD caref ully, you will notice that ABC is a right angle triangle.Where AB, BC and AC is the breath, length and diagonal of the rectangle.Now in right angle triangle ABC

AC2 = AB2 + BC2

⇒ (4b)2 = b2 + l2

⇒ 16b2 - b2 = l2

⇒ (16 - 1)b2 = l2

⇒ 15b2 = l2

⇒ l2 = 15b2

⇒ l2

b2 =

15

1

⇒ ( l

b )2 =

15

1

⇒ l

b =

√15

1

⇒ l:b = √15:1

Step 4

Theref ore we can say that the ratio of length and breath of the rectangle is √15:1.



(13) 12.6

Step 1

Lets assume r is the radius of the circle. The circumf erence of the circle = 2πr

Step 2

According to question the circumf erence of a circle is 66.6 cm more than its radius.Theref ore 2πr = r + 66.6⇒ 2πr - r = 66.6⇒ r(2π - 1) = 66.6

⇒ r(2 x 22

7 - 1) = 66.6

⇒ r( 37

7 ) = 66.6

⇒ r = 66.6 x 7

37

⇒ r = 12.6

Step 3

Now the radius of the circle is 12.6 cm.



(14) 35

Step 1

Lets assume r and d are the radius and diameter of the semi-circular shape respectively.Diameter of the semi-circular shape = 2r,Perimeter of the semi-circular shape = πr + d= πr + 2r= r(π + 2)

= r( 22

7 + 2)

= r( 22

7 + 2)

= r( 22 + 14

7 )

= 36r

7

Step 2

According to question perimeter of the semi-circular shape is 180 cm.

Theref ore 36r

7 = 180

⇒ r = 180 × 7

36

⇒ r = 35

Step 3

Now radius of the semi-circular shape = 35 cm



(15) 18

Step 1

We know that the area of rectangle = length × breath,Area of rectangle ABCD = AB × AD,Area of rectangle AEFG = AE × AG

Step 2

If you look at the question caref ully, you will notice thatArea of rectangle ABCD = Area of rectangle AEFG⇒ AB × AD = AE × AG -----(1)If you look at the image, you will notice thatAB = AE + EB,AG = AD + DGtheref ore (1) becomes (AE + EB)AD = AE(AD + DG)⇒ (AE × AD) + (EB × AD) = (AE × AD) + (AE × DG)⇒ (EB × AD) = (AE × AD) - (AE × AD) + (AE × DG)⇒ (EB × AD) = 0 + (AE × DG)⇒ EB × AD = AE × DG

⇒ AD = AE × DG

EB

Since AE=36 cm, DG=6 cm and EB=12 cm,

Now AD = 36 × 6

12

= 216

12

= 18

Step 3

Theref ore the length of AD = 18 cm.



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